Solving Polynomials: Hints, Techniques & Solutions

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The discussion centers around solving a complex trigonometric equation involving sine and cosine functions, which the original poster finds challenging. Participants agree that an analytic solution is unlikely, suggesting numerical methods like the Bisection or Newton's method as more viable options. One user attempts to simplify the equation but concludes that the simplification may not be particularly useful. The conversation highlights a misunderstanding regarding the classification of the equation as a polynomial, with users clarifying that it is not. Overall, the consensus leans towards using numerical techniques for solving the equation effectively.
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Homework Statement



Solve for x,

225*sin(x)/x^6-225*cos(x)/x^5-90*sin(x)/x^4+15*cos(x)/x^3-5/(2*x^3)=0


Homework Equations



Finding this very complicated to solve, are there any useful hints or techniques we should know about?


The Attempt at a Solution



Have used numerical method using mathematics software and plotted a graph to identify where the function crosses the x-axis. Would prefer a more analytic approach.

Thank you in advance.
 
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ts547 said:

Homework Statement



Solve for x,

225*sin(x)/x^6-225*cos(x)/x^5-90*sin(x)/x^4+15*cos(x)/x^3-5/(2*x^3)=0


Homework Equations



Finding this very complicated to solve, are there any useful hints or techniques we should know about?


The Attempt at a Solution



Have used numerical method using mathematics software and plotted a graph to identify where the function crosses the x-axis. Would prefer a more analytic approach.

Thank you in advance.

I think that you are out of luck regarding an analytic solution.

BTW, is this your equation?
225\frac{sin(x)}{x^6} - 225 \frac{cos(x)}{x^5} - 90\frac{sin(x)}{x^4} + 15\frac{cos(x)}{x^3} - \frac{5}{2x^3} = 0
 
Yeh that's it. I haven't learned how to do the fancy writing yet. I didnt think there would be an easy way of doing this.
 
Unless there's some funny trick to recognize here, there's no way to solve this algebraically. It's best to use a numerical technique. i.e. Bisection method, Newton's method, etc.
 
ts547 said:
Yeh that's it. I haven't learned how to do the fancy writing yet. I didnt think there would be an easy way of doing this.
You can see the LaTeX I wrote by clicking the equation.
 
Heck, I managed to simplify it this equation (check work?):

<br /> (\frac{15}{x^3} - \frac{6}{x})sin(x) + (1 - \frac{15}{x^2})cos(x) = \frac{1}{6}<br />

Edit: LaTeX isn't the easiest, heh. Also, I'm not sure that simplification is even very useful.
 
I hope that you are aware that this is not a matter of "solving polynomials"! The equation you give is not a polynomial.
 
Apphysicist - Haha good simplification. Not very useful I don't think. :) Never mind ill stick with the numerical approach.

HallsofIvy - Ok no its not a polynomial. Didnt know what else to call it at the time. If your so clever help me with this

https://www.physicsforums.com/showthread.php?p=3075732#post3075732

then you can point out technicalities all you want.
 
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